# Log Equation

#### alternative

Hello, Could you help me solve this ?

I'd appreciate ur help, thanks a lot

#### Prove It

MHF Helper
Hello, Could you help me solve this ?

I'd appreciate ur help, thanks a lot
$$\displaystyle \frac{\log_2{(x^2 + 2x - 7)}}{\log_2{(x - 3)}} = 1$$

$$\displaystyle \log_2{(x^2 + 2x - 7)} = \log_2{(x - 3)}$$

$$\displaystyle x^2 + 2x - 7 = x - 3$$

$$\displaystyle x^2 + x - 4 = 0$$

$$\displaystyle x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 4 = 0$$

$$\displaystyle \left(x + \frac{1}{2}\right)^2 - \frac{17}{4} = 0$$

$$\displaystyle \left(x + \frac{1}{2}\right)^2 = \frac{17}{4}$$

$$\displaystyle x + \frac{1}{2}= \pm \frac{\sqrt{17}}{2}$$

$$\displaystyle x = -\frac{1}{2} \pm \frac{\sqrt{17}}{2}$$.

Therefore

$$\displaystyle x = \frac{-1 - \sqrt{17}}{2}$$ or $$\displaystyle x = \frac{-1 + \sqrt{17}}{2}$$.

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